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Expert-verified Found in: Page 172 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Solve the given initial value problem ${\mathbf{y}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\text{'}}{\mathbf{+}}{\mathbf{17}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}$${\mathbf{y}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{1}}{\mathbf{,}}$${\mathbf{y}}{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$.

The solution of the given initial value $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+17\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{cos}4\mathrm{t}\right)$ when $\mathrm{y}\left(0\right)=1$ and $\mathrm{y}\text{'}\left(0\right)=-1.$

See the step by step solution

## Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots ${\alpha }{±}{i}{\beta }$ , then the general solution is given as:

${y}\left(t\right){=}{{c}}_{{1}}{{e}}^{\alpha t}{c}{o}{s}{\beta }{t}{+}{{c}}_{{2}}{{e}}^{\alpha t}{s}{i}{n}{\beta }{t}$.

## Step 2: Finding the roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+17\mathrm{y}=0$

Then the auxiliary equation ${\mathrm{r}}^{2}+2\mathrm{r}+17=0$

Solve the auxiliary equation to obtain the roots.

$\begin{array}{c}\mathrm{r}=\frac{-2±\sqrt{{2}^{2}-4×1×17}}{2×1}\\ \mathrm{r}=\frac{-2±\sqrt{4-68}}{}\\ \mathrm{r}=\frac{-2±\sqrt{-64}}{}\\ \mathrm{r}=\frac{-2±8\mathrm{i}}{}\\ \mathrm{r}=-1±4\mathrm{i}\end{array}$

Therefore, the general solution is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-1×\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(4\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(4\mathrm{t}\right)\right)\\ ={\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(4\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(4\mathrm{t}\right)\right)\end{array}$

## Step 3: Finding the values of C1 and C2

Given initial conditions are $\mathrm{y}\left(0\right)=1$ and $\mathrm{y}\text{'}\left(0\right)=-1.$

role="math" localid="1654841671923" $\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{e}}^{-0}\left({\mathrm{c}}_{1}\mathrm{cos}\left(4×0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(4×0\right)\right)\\ {\mathrm{c}}_{1}=1\end{array}$

And

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-{\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}4\mathrm{t}+{\mathrm{c}}_{2}\mathrm{sin}4\mathrm{t}\right)+{\mathrm{e}}^{-\mathrm{t}}\left(-4{\mathrm{c}}_{1}\mathrm{sint}+4{\mathrm{c}}_{2}\mathrm{cost}\right)$

Then,

$\begin{array}{l}\mathrm{y}\text{'}\left(0\right)=-{\mathrm{e}}^{-0}\left({\mathrm{c}}_{1}\mathrm{cos}\left(4×0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(4×0\right)\right)+{\mathrm{e}}^{-0}\left(-4{\mathrm{c}}_{1}\mathrm{sin}\left(4×0\right)+4{\mathrm{c}}_{2}\mathrm{cos}\left(4×0\right)\right)\\ -{\mathrm{c}}_{1}+4{\mathrm{c}}_{2}=-1\end{array}$

Substitute ${c}_{1}$ in the above equation

$\begin{array}{c}-1+{\mathrm{c}}_{2}=-1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{2}=0\end{array}$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{cos}4\mathrm{t}\right)$. ### Want to see more solutions like these? 